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Assoli18 [71]
3 years ago
6

Find the equation of the line that passes through the given points, in Slope-Intercept Form. (-2, -7) and (3, 8)

Mathematics
1 answer:
julia-pushkina [17]3 years ago
7 0

Answer:

y=3x-1

Step-by-step explanation:

m=y2-y1/x2-x1

m=8-(-7)/3-(-2)

m=8+7/3+2

m=15/5

m=3

y-y1=m(x-x1)

y-(-7)=3(x-(-2)

y+7=3(x+2)

y+7=3x+6

y=3x+6-7

y=3x-1

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Two cars start out together from the same place. they travel in opposite​ directions, with one of them traveling 5 miles per hou
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<u>Annotation</u>
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.


One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5  (first equation)

The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)

Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262

Plug the v₁ as  (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.

Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.

Answer
\boxed{\boxed{ v_{1}=68mph} }
\boxed{\boxed{ v_{2}=63mph} }
7 0
4 years ago
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